Question: Simplify the following expression: $y = \dfrac{-2x^2+1x+10}{-2x + 5}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-2)}{(10)} &=& -20 \\ {a} + {b} &=& &=& {1} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-20$ and add them together. Remember, since $-20$ is negative, one of the factors must be negative. The factors that add up to ${1}$ will be your ${a}$ and ${b}$ When ${a}$ is ${5}$ and ${b}$ is ${-4}$ $ \begin{eqnarray} {ab} &=& ({5})({-4}) &=& -20 \\ {a} + {b} &=& {5} + {-4} &=& 1 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-2}x^2 +{5}x) + ({-4}x +{10}) $ Factor out the common factors: $ x(-2x + 5) + 2(-2x + 5)$ Now factor out $(-2x + 5)$ $ (-2x + 5)(x + 2)$ The original expression can therefore be written: $ \dfrac{(-2x + 5)(x + 2)}{-2x + 5}$ We are dividing by $-2x + 5$ , so $-2x + 5 \neq 0$ Therefore, $x \neq \frac{5}{2}$ This leaves us with $x + 2; x \neq \frac{5}{2}$.